I bought a new calculator at the Dollar Store and, sure enough, it’s defective. After some trial and error with it, I discovered that each digit in the display contained the same two pairs of elements (out of the seven elements labeled A to G below) that were somehow ‘cross-wired’. That is, if one element was called upon to illuminate, its partner would illuminate instead. If both were supposed to illuminate, neither would! For example, if A/D and B/F were the faulty pairs, the number 3 would simply display as F/G/C, as illustrated below.

A

F

B

G

E

C

D

Based on the illuminated elements for each digit given below, find the faulty pairs to then solve the following 3-digit by 2-digit multiplication:

According to the puzzle, "each digit in the display contained the same two pairs of elements that were somehow ‘cross-wired’."

In the proposed solution: 153x 497497

If 1, normally represented by B/C shows as C/D, B must be crossed with D and the other crossed pair not involved with 1.

Then 5 changes from AFGCD to FBGEC, showing F swapped with E.

So the two swaps are F with E and B with D.

A 3, normally ABCDG, would, under this pair of swaps become ACG, but the puzzle shows GEC. Why E, rather than A? E and F are not involved with a 3 while B and D are both present, resulting in their extinguishment.

Then, a 4, normally FGBC should change to EGDC, but shows on the puzzle table as FGCD, with its F unchanged rather than swapped to the E.

The 9 is normally either ABFGC or ABFGCD. If the former, it would change to ADEGC and if the latter, AEGC, while the table shows neither.

The 7, normally ABC would change to ADC, rather than FECD.

Now, in case 1 were normally FE, rather than the BC I had assumed above, then the 1 as CD would indicate either F changed with C and E with D or F with D and E with C. Either would leave the 3 with ABFEG, rather than GEC.